AP Unit 8 Solutions

1. Compute ∫₀³ x²dx=9 and divide by 3−0=3. Reason: Average value is integral divided by interval length.
2. Solve c²=3 and retain c=√3 in [0,3]. Reason: The point must lie in the specified interval.

Answer: Average 3; c=√3.

1. Approximate ∫Tdt by ½(10+14)·2+½(14+20)·3=75 °C·min. Reason: Use the nonuniform widths and average endpoint temperatures.
2. Divide by 5 min to get 15°C. Reason: Average value normalizes by the full interval length.

Answer: 15°C.

1. Displacement ∫₀³vdt=[t²−2t]₀³=3 m, so final position is 5+3=8 m. Reason: Position is initial position plus velocity accumulation.
2. Velocity changes sign at t=1; distance=−∫₀¹v+∫₁³v=1+4=5 m. Reason: Speed is the absolute value of velocity.

Answer: s(3)=8 m; distance 5 m.

1. Integrate to v(t)=3t²−2. Reason: Include the initial velocity in the acceleration integral.
2. At t=2, v=12−2=10. Solve 3t²−2=0 for t≥0. Reason: A direction change requires velocity zero with a sign change, which this increasing velocity has.

Answer: v(2)=10 m/s; direction changes at t=√(2/3) s.

1. Signed areas are ½·2·4=4 and −½·3·2=−3. Reason: Velocity area represents displacement with signs.
2. Add signed areas for 1 m and their magnitudes for 7 m. Reason: Distance integrates speed.

Answer: Displacement 1 m; distance 7 m.

1. V(t)=10+∫₀ᵗ(6−2u)du=10+6t−t². Reason: Integrate net rate and include the initial amount.
2. Critical time is 3; compare V(0)=10,V(3)=19,V(4)=18. Reason: A maximum on a closed interval occurs among endpoints and derivative-zero candidates.

Answer: 19 L at t=3 min.

1. Net rate is 5t−2. Reason: Subtract outflow from inflow.
2. Net change is ∫₀³(5t−2)dt=22.5−6=16.5. Reason: Integrate the net rate over the time interval.
3. Add 50 to obtain 66.5. Reason: The integral gives change rather than final inventory; the model treats inventory continuously.

Answer: 66.5 items in the continuous model.

1. Rate units times day units produce dollars. Reason: Integral accumulation multiplies rate by elapsed time.
2. Add the change to the starting balance. Reason: The net-change principle connects initial and final amounts.

Answer: Net balance change from day 2 to day 7, in dollars; final balance is 100+∫₂⁷ r(t)dt.

1. Intersections solve x=x², giving 0 and 1. Reason: These are the region’s bounds.
2. On (0,1), x>x², so area=∫₀¹(x−x²)dx=1/2−1/3=1/6. Reason: Integrate top minus bottom.

Answer: 1/6 square unit.

1. Solve x²=4 to find −2 and 2; the constant curve is above. Reason: Intersections and ordering determine vertical strips.
2. Evaluate ∫₋₂²(4−x²)dx=16−16/3=32/3. Reason: Area is the integral of positive strip height.

Answer: 32/3 square units.

1. Intersections are y=0 and y=1; y is right of y² between them. Reason: Horizontal strips use right minus left.
2. Integrate ∫₀¹(y−y²)dy=1/6. Reason: Use y-bounds and dy consistently.

Answer: 1/6 square unit.

1. A horizontal strip extends from x=0 to x=y². Reason: Its width is right minus left.
2. Area=∫₀² y²dy=8/3. Reason: Integrate the horizontal strip width in y.

Answer: 8/3 square units.

1. Crossings occur at −1,0,1. The cubic is above on [−1,0] and the line above on [0,1]. Reason: Order changes at the interior crossing.
2. Use symmetry: area=2∫₀¹(x−x³)dx=2(1/2−1/4)=1/2. Reason: Both lobes have equal positive area.

Answer: 1/2 square unit.

1. Sine is positive on (0,π) and negative on (π,2π). Reason: Split at the sign change.
2. The two lobe magnitudes are 2 each; signed values are +2 and −2. Reason: Geometric area adds magnitudes while the integral retains signs.

Answer: Area 4; signed integral 0.

1. The side is s=x−x², so A=(x−x²)². Reason: The base segment supplies the square’s side.
2. Integrate ∫₀¹(x²−2x³+x⁴)dx=1/3−1/2+1/5=1/30. Reason: Volume integrates cross-sectional area.

Answer: 1/30 cubic unit.

1. Area is 3(2−x)². Reason: Rectangle area is base times height.
2. V=3∫₀²(2−x)²dx=3·8/3=8. Reason: Integrate the known slice area.

Answer: 8 cubic units.

1. For side s=x−x², triangle area is √3·s²/4. Reason: The equilateral-triangle geometry specifies height.
2. V=(√3/4)∫₀¹(x−x²)²dx=(√3/4)(1/30). Reason: Use the same base-strip length in the triangle area formula.

Answer: √3/120 cubic unit.

1. Radius=(2−x)/2, giving A=π(2−x)²/8. Reason: A semicircle is half a disc; the given segment is diameter.
2. Integrate V=(π/8)(8/3)=π/3. Reason: Volume is the integral of slice area.

Answer: π/3 cubic units.

1. The section area is π(√x)²=πx. Reason: Use the full circle area because these sections are circles.
2. V=π∫₀²x dx=2π. Reason: Integrate the known area function.

Answer: 2π cubic units.

1. A perpendicular slice forms a disc of radius √x. Reason: The region reaches the x-axis.
2. V=π∫₀⁴x dx=8π. Reason: Square the radius to get slice area.

Answer: 8π cubic units.

1. Use horizontal slices with radius √y. Reason: Discs must be perpendicular to the vertical axis.
2. V=π∫₀⁴y dy=8π. Reason: Use the y-bounds with dy.

Answer: 8π cubic units.

1. Radius is (1+x)−1=x; the region reaches the axis. Reason: Distance from the shifted axis determines the disc radius.
2. V=π∫₀²x²dx=8π/3. Reason: Integrate the disc areas.

Answer: 8π/3 cubic units.

1. Use horizontal slices with radius (y−1)−(−1)=y. Reason: The axis is vertical and the region reaches it.
2. V=π∫₀²y²dy=8π/3. Reason: The disc formula uses the radial distance squared.

Answer: 8π/3 cubic units.

1. Outer radius is 2, inner radius is x. Reason: Both are distances from y=0, with a hole for x>0.
2. V=π∫₀²(4−x²)dx=π(8−8/3)=16π/3. Reason: Subtract inner disc area from outer disc area.

Answer: 16π/3 cubic units.

1. Horizontal slices have outer radius y and inner radius y². Reason: The right boundary is farther from x=0.
2. V=π∫₀¹(y²−y⁴)dy=π(1/3−1/5)=2π/15. Reason: Use washer areas with dy.

Answer: 2π/15 cubic unit.

1. Outer radius is 3−x and inner radius is 1. Reason: The lower boundary is farther from the axis above the region.
2. V=π∫₀²[(3−x)²−1]dx=π∫₀²(8−6x+x²)dx=20π/3. Reason: Square each distance separately, then subtract.

Answer: 20π/3 cubic units.

1. Outer radius is y+1 and inner radius is y²+1. Reason: Both distances start at the same shifted vertical axis.
2. V=π∫₀¹[(y+1)²−(y²+1)²]dy=π∫₀¹(2y−y²−y⁴)dy. Reason: Use horizontal washers.
3. The value is π(1−1/3−1/5)=7π/15. Reason: Evaluate the polynomial antiderivative.

Answer: 7π/15 cubic unit.

1. Solve e^(−c²)=c in [0,1], giving c≈0.6529186404. Reason: The curves exchange order at their unique crossing; e^(−x²)−x is strictly decreasing for x≥0.
2. Use ∫₀ᶜ[e^(−x²)−x]dx+∫꜀¹[x−e^(−x²)]dx. Reason: Area needs positive height on both pieces.
3. Numerically evaluate the sum, approximately 0.4686590006, and round to 0.469. Reason: Keep the unrounded root and integrals in calculator memory.

Answer: Intersection c≈0.652919; area ≈0.469 square unit.

1. Outer disc area is 9π; inner hole area is π. Reason: Each radius determines a separate disc area.
2. Subtract to obtain 8π. Reason: A washer is outer area minus the hole, not a disc with radius 2.

Answer: 8π cm²; the student’s 4π cm² squares the ring’s thickness instead of subtracting disc areas.

1. Triangle area is ½·s·2s=s². Reason: Use the stated base-height relationship.
2. Semicircle radius is s/2, so its area is ½π(s/2)²=πs²/8. Reason: The supplied segment is diameter, not radius.
3. Circle area is πs². Reason: Here the supplied segment is explicitly the radius.

Answer: Triangle: s²; semicircle: πs²/8; circle: πs².